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Mathematics 8 Online
OpenStudy (anonymous):

medals will be given which graph represents the solution to the given system? y=-2x + 5 and y= 3x - 2

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

choose the right graph

OpenStudy (the_fizicx99):

This is very simple, but I'm on my phone, need paper to do this :/ sry

OpenStudy (anonymous):

@tHe_FiZiCx99 YESS I NEED HELPP

OpenStudy (the_fizicx99):

I could use a graphing calculator but my browser would crash lol >.> @kelliegirl33 is better at explaining in detail than I am :)

OpenStudy (anonymous):

@kelliegirl33 can you help please : (

OpenStudy (anonymous):

It is the last one....because when you sub in (-7,19) into both equations, the points satisfy both equations.

OpenStudy (anonymous):

thank youuuu i have a few more questions can you help with ?

OpenStudy (anonymous):

I can try

OpenStudy (anonymous):

which graph represents the solution to the given system? -4x + 2y=-11 and -2x + 2y =-15

OpenStudy (anonymous):

-4x + 2y = -11 2y = 4x - 11 y = 2x - 11/2 (the y intercept is -11/2 which equals - 5 1/2 and the slope is 2) -2x + 2y = -15 2y = 2x - 15 y = x - 15/2 (y intercept is -15/2 which equals - 7 1/2 and slope is 1) last graph ( I think)

OpenStudy (anonymous):

thank you what about this one how many solutions does the system have ? 2x=-10y+6 and x+5y=3 A.one B.two C.infinitely many D.none

OpenStudy (anonymous):

LAST ONE how many solutions does the system of equations have ? y=6x + 2 and 3y-18x = 12 one two infintely many noe

OpenStudy (anonymous):

2x = -10y + 6 change to 2x + 10y = 6 which reduces to x + 5y = 3. This is the same as the other equation, therefore there is INFINITE SOLUTIONS ============================================ y = 6x + 2 (sub 6x + 2 in for y in the other equation) 3y - 18x = 12 3(6x + 2) - 18x = 12 18x + 6 - 18x = 12 18x - 18x = 12 - 6 0 = -6 (incorrect) This one has NO SOLUTIONS

OpenStudy (anonymous):

OMGGGGGG THANK YOU SO MUCH FOR YOUR HELPPPP !!!! i really appreciate it so much

OpenStudy (anonymous):

no problem...glad to help :)

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